I want to run an outside light and receptacle about 250 feet from the main panel load center. It I use a 15 amp circuit breaker, is 10 gauge wire large enough?

As long as the load does not exceed 6 amps, you will be fine. Any load higher than 6 amps will beging to increase the voltage drop above the recommmended maximum 3%.

I'm doing something very similar. I'm running a 300 ft circuit for some lighting at the back of my church.
The load will be only about 8 Amps max, but I'm running 10 AWG because in the case of a fault at the very end of the circuit, I want minimal clearing time. I'm not that worried about the voltage drop.

What is the load on your new circuit? Look at some of the voltage drop calculators online to see if 10 AWG this meets your requirements.
I'm using 10 AWG because I want to be safe, not sorry. It's worth the extra money. Again, I'm more worried about the fault current than the voltage drop.
Hope this helps!
Neal….

Neal, what does the wire size have to do with "fault clearing time"?

To: stanfortyman
My thinking is the lower resistance of the 10 AWG wire over this long run would allow more current available to the 15 Amp CB so it will operate faster, thus reduce the I2t thermal damage to the wire.

In the past, I've seen some long branch circuits (usually underground circuits to signs) that shorted and the CB would buzz loudly (and get hot) for a few seconds before clearing the fault. Yes, the 12 AWG conductor to the sign was enough to supply the 4 Amps to the sign, but the excessive resistance of the wire limited the fault current to a level that allowed a LOT of high current through the circuit that it damaged the conductors and I had to replace them.

It's been my experience that the lower the resistance of the brance circuit. The faster the breaker tripps (clears) the faualt.

Does this make sense?
Thanks,
Neal...

Quote:

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Originally Posted by NealSellers

It's been my experience that the lower the resistance of the brance circuit. the faster the breaker tripps (clears) the faualt.

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I've never experienced this.

Neal, when designing a circuit, voltage drop is the initial issue to deal with. The load must have ample voltage to operate properly.

Your theory of low impedance (ac resistance) of a branch circuit conductors is correct, by using a wire size large enough, however, is a secondary result of using the proper size wire for the load, distance and reducing voltage drop. All simple Ohm's Law.

For your example of 8 amps 120 volts at 300 feet, #8 wire is needed, so your not following your own theory.

Increasing the size of the equipment grounding conductor proportionally to the branch circuit conductors when increased to reduce voltage drop is required to reduce the impedance of the equipment grounding conductor and the branch circuit to allow a circuit breaker to open properly during a fault condition.

This, of course, will protect the entire circuit from damage due to I2R thermal indirectly.


In your example of 4 amps on #12 wire, distance unknown, I assume the distance was great and the #12 wire was too small, the wire becomes another "load" in the circuit.

Your thinking is correct, however, for the incorrect reason. A circuit must have the proper voltage to work first, lower impedance of the circuit is a secondary by-product.

Fault clearing time is dealt with by proper selection of overcurrent protection devices. Proper circuit design should always include the OCPD and choosing the proper characteristics of fault time curves.